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Let $(E,d)$ be a metric space, $(\mu_t)_{t\in I}$ be a net of finite nonnegative measure on $\mathcal B(E)$ and $\mu$ be a finite nonnegative measure on $\mathcal B(E)$ such that $(\mu_t)_{t\in I}\to\mu$ weakly, i.e. $$\left(\int f\:{\rm d}\mu_t\right)_{t\in I}\to\int f\:{\rm d}\mu\tag1.$$

By the Portmanteau theorem, $(1)$ implies (it is even equivalent to) $$\lim_{t\in I}\mu_t(B)=\mu(B)\;\;\;\text{for all }B\in\mathcal B(E)\text{ with }\mu(\partial B)=0\tag2.$$

Are we able to show an analogous result for finite signed measures: If $(\nu_t)_{t\in I}$ is a net of finite signed measure on $\mathcal B(E)$ and $\nu$ be a finite signed measure on $\mathcal B(E)$ such that $(\nu_t)_{t\in I}\to\nu$ weakly, does it follow that $$\lim_{t\in I}|\nu_t-\nu|(B)=0\;\;\;\text{for all }B\in\mathcal B(E)\text{ with }\mu(\partial B)=0\tag3$$ holds?

0xbadf00d
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1 Answers1

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The example I quoted for your previous question works here too. On $[0,1]$ let $\nu_n=\delta_{\frac 1 2+\frac1 n}-\delta_{\frac 1 2-\frac1 n}$ and $\nu=0$. Take $B=[0,1]$.

Kavi Rama Murthy
  • 311,013
  • While $(3)$ might fail to hold, it would immediately follow from the identity $(1)$ I'm asking for in this question that $(\mu_t){t\in I}\to\mu$ weakly implies $$\liminf{t\in I}|\mu_t|(C)\ge|\mu|(A);;;\text{for all open }C\subseteq E\tag4$$ and I guess we are able to show a similar result for closed sets and the $\limsup$ as well. Maybe you can take a look. – 0xbadf00d Dec 31 '20 at 12:02